Question

A 47 gram golf ball is driven from the tee with an initial speed of 52 m/sec and rises to a height of 24.6 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8 m below its highest pint

Answer 1

Answer:

a) 52.2 J

b) 48.77 m/s

Explanation:

a)47 g = 0.047 jg

The kinetic (and total mechanical energy) of the ball at the ground is

$E = mv^2/2 = 63.544 J$

The potential energy of the ball at its highest point 24.6m is. Let g = 9.8m/s2

$E_h = mgh = 0.047*9.8*24.6 = 11.33 J$

Since the potential energy at the highest height is less than the total mechanical energy on ground, the difference must be kinetic energy

$E_k = E - E_h = 63.544 - 11.33 = 52.2J$

b) 8m below 24.6m is 16.6m. The potential energy at this point is

$E_{p8} = mgh = 0.047*9.8*16.6 = 7.64 J$

And so the kinetic energy at this point is

$E_{k8} = E - E_{p8} = 63.544 - 7.64 = 55.9 J$

So the speed is

$mv^2/2 = 55.9$

$v^2 = 2*55.9/0.047 = 2378.64$

$v = \sqrt{2378.64} = 48.77 m/s$